Question 1 of 6: Stability limit, phase margin and steady-state errors of a non-minimum-phase loop

Question 1: Stability limit, phase margin and steady-state errors of a non-minimum-phase loop (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A unity-feedback loop around a third-order, non-minimum-phase plant driven by a pure gain, with a disturbance entering at the plant output.

Find. The gain at which the loop first sustains oscillation and that oscillation frequency; the phase margin at half that gain; the steady-state tracking error to a unit ramp with a unit output disturbance; and the steady-state tracking error to $r(t)=4\sin t$.

Approach. Routh’s criterion on the closed-loop characteristic polynomial fixes the stability limit and, through its auxiliary polynomial, the oscillation frequency; the mirror-image zero–pole pair then gives a closed-form gain crossover for the phase margin; and the single error transfer function $E=(R+D)/(1+L)$ answers both steady-state parts.

1. Part (a) — write the closed-loop characteristic polynomial. Sustained oscillation means a pair of closed-loop poles sits exactly on the imaginary axis, so start from $1+C(s)P(s)=0$ cleared of denominators:$$s(2+s)(1+s)+6K(1-s)=0 \;\Longrightarrow\; s^{3}+3s^{2}+(2-6K)s+6K=0$$The right-half-plane zero at $s=+1$ is what makes the $s$-coefficient shrink as $K$ grows — that is the mechanism that destabilises this loop.
2. Apply the Routh criterion. With $a_{3}=1$, $a_{2}=3$, $a_{1}=2-6K$, $a_{0}=6K$, the array is$$\begin{array}{c|cc} s^{3} & 1 & 2-6K\\ s^{2} & 3 & 6K\\ s^{1} & \dfrac{3(2-6K)-6K}{3}=2-8K & 0\\ s^{0} & 6K & \end{array}$$Both surviving first-column entries must stay positive, giving $6K\gt 0$ and $2-8K\gt 0$, i.e. the loop is stable only on $0\lt K\lt 0.25$.
3. Read the stability limit off the vanishing Routh entry. The $s^{1}$ entry is the one that reaches zero first, and it does so at$$\boxed{K_{max}=\tfrac{1}{4}=0.250}$$At that gain the $s^{2}$ row becomes an auxiliary polynomial whose roots are the imaginary-axis poles: $3s^{2}+6K_{max}=3s^{2}+1.5=0$, so $s^{2}=-0.5$ and
4. Convert the auxiliary root to the oscillation frequency. $$\omega_{osc}=\sqrt{0.5}=\boxed{0.7071\ \text{rad/s}}$$Equivalently $\omega_{osc}^{2}=p/(p+2)$ with $p=2$ — note this frequency does not depend on the numerator gain 6 at all, only on the pole locations. A period of $2\pi/\omega_{osc}=8.886$ s is what an oscilloscope would show at $K=0.25$.
5. Part (b) — simplify the loop magnitude at $K=K_{max}/2=0.125$. Writing $L(j\omega)=6K(1-j\omega)\big/\!\left[j\omega(2+j\omega)(1+j\omega)\right]$, the factor $\sqrt{1+\omega^{2}}$ contributed by the right-half-plane zero at $+1$ is identical to the one contributed by the pole at $-1$, so the two cancel exactly:$$|L(j\omega)|=\frac{6K\sqrt{1+\omega^{2}}}{\omega\sqrt{4+\omega^{2}}\,\sqrt{1+\omega^{2}}}=\frac{6K}{\omega\sqrt{4+\omega^{2}}}$$That mirror-image pairing is why the gain crossover has a closed form here even though the phase does not.
6. Solve for the gain crossover. Setting $|L|=1$ and squaring gives a quadratic in $\omega^{2}$:$$\omega^{4}+4\omega^{2}-(6K)^{2}=0 \;\Longrightarrow\; \omega_{gc}^{2}=\frac{-4+\sqrt{16+4(0.75)^{2}}}{2}=0.13600$$so $\omega_{gc}=0.3688$ rad/s. Solving numerically for $|L|=1$ reproduces the same value to four figures, which is the check worth doing before spending marks on the phase.
7. Evaluate the phase there and take the margin. The phase is the sum of four contributions — the RHP zero subtracts $\arctan\omega$ rather than adding it:$$\angle L=-\arctan\omega_{gc}-90^{\circ}-\arctan\tfrac{\omega_{gc}}{2}-\arctan\omega_{gc}=-20.24^{\circ}-90^{\circ}-10.45^{\circ}-20.24^{\circ}=-140.93^{\circ}$$$$\text{PM}=180^{\circ}+\angle L(j\omega_{gc})=\boxed{39.1^{\circ}}$$Halving the gain from its stability limit therefore buys about 39 degrees of phase margin, a respectable but not generous figure for a plant with a right-half-plane zero.
8. Part (c) — build the error transfer function with the disturbance where the figure puts it. The block diagram subtracts $d$ at the plant OUTPUT, so $y=P\,u-d$ and $e=r-y$ give$$E(s)=\frac{R(s)+D(s)}{1+L(s)},\qquad L(s)=C(s)P(s)$$Reading that sign off the drawing matters: had $d$ entered at the plant input it would be weighted by $P(0)$ and the answer below would change.
9. Take the two steady-state limits separately. For the unit-slope ramp $R=1/s^{2}$ the velocity constant is $K_{v}=\lim_{s\to 0}sL(s)=6K/2=3K=0.375$, so$$e_{ramp}(\infty)=\frac{1}{K_{v}}=\frac{1}{3K}=\boxed{2.667}$$For the constant disturbance $D=1/s$ the limit is $\lim_{s\to 0}\,s\,(1/s)/(1+L(s))=0$, because $L$ has a pole at the origin and $1+L$ blows up. A constant disturbance entering at the output is therefore rejected completely, and the total steady-state tracking error is $2.667$ — the ramp alone.
10. Part (d) — a sinusoidal command is a frequency-response question, not a limit. With $d=0$ the error is $E=S\,R$ with the sensitivity $S=1/(1+L)$, and the steady-state error is the sinusoid $r$ scaled and shifted by $S$ evaluated at the command frequency $\omega=1$ rad/s. Substituting $L(j1)=0.75(1-j)/(j(2+j)(1+j))=-0.300+j0.150$:$$S(j1)=\frac{1}{1+L(j1)}=\frac{1}{0.700+j0.150},\qquad |S(j1)|=1.397,\quad \angle S(j1)=-12.09^{\circ}$$
11. Write the answer as a time function. $$e(t)=4\,|S(j1)|\sin\!\left(t+\angle S(j1)\right)=\boxed{5.587\sin\!\left(t-12.09^{\circ}\right)}$$Because $|S(j1)|\gt 1$, the steady-state error is larger than the 4-unit command that produced it. That is not an arithmetic slip: 1 rad/s is above the loop bandwidth and close to the sensitivity peak forced by the right-half-plane zero, so the loop amplifies rather than attenuates the tracking error there.